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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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If the points $(0,4)$ and $(0,2)$ are respectively the vertex and focus of a parabola,then find the equation of the parabola.

$\begin{array}{1 1}x^2+8y-32=0\\x^2-8y+32=0\\x^2+4y+32=0\\x^2-4y+32=0\end{array} $

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  • General equation of the parabola whose vertex are $(h,k)$ and open downward is $(x-h)^2=-4a(y-k)$
Answer : $x^2+8y-32=0$
The coordinates of the vertex is $(0,4)$
The coordinates of the focus is $(0,2)$
It is clear that the vertex and the focus lies on the positive side of the y-axis.
Hence the curve is open downwards.
The equation of the form $(x-h)^2=4a(y-k)$
(ie) $(x-0)^2=-4\times 2(y-4)$
On simplifying we get,
$x^2=-8(y-4)$
$\Rightarrow x^2=-8y+32$
$x^2+8y-32$ is the required equation of the parabola.
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